Suppose that is a permutation with 8 cycles, all of whose lengths divide 12. If has 13 cycles, has 16, has 15, has 25, and has 27, what is the cycle type of w?

What can we say if we drop the guarantee that all cycle lengths divide 12?

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Hm, say there are A 1s, B 2s, C 3s, D 4s, E 6s, F 12s.

A+B+C+D+E+F = 8

A+2B+C+2D+2E+2F = 13

A+B+3C+D+3E+3F = 16

A+2B+C+4D+2E+4F = 15

A+2B+3C+2D+6E+6F = 25

A+2B+3C+4E+6E+12F = 27

This has a unique solution (1,1,2,2,3,-1). What did I do wrong?

I bet row-reducing this matrix amounts to computing the standard Mobius function…

By:

Allen Knutsonon August 6, 2015at 5:12 AM

Just arithmetic, probably: your proposed solution doesn’t satisfy the last equation.

I had thought it was going to be just the usual Moebius function (or a simple twist), but it seems a little bit more complicated. I believe you would get mu if I gave you data on number of fixed points instead of number of cycles.

By:

JBLon August 6, 2015at 9:07 AM

Oh, I understand your last comment now. Yes, that’s exactly what’s going on; the matrix in question is the product of the zeta matrix for divisibility relation, a diagonal matrix, and the transpose of the zeta matrix, so when you row-reduce you’re exactly doing some Möbius inversion

By:

JBLon August 7, 2015at 8:17 PM